Difference between revisions of "1968 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | |
| + | Let the number of votes for the bill the first time be <math>a</math> and the number of votes for the bill the second time be <math>b</math>. Then, from the problem, we know that <math>b=\frac{12}{11}(400-a)</math>. Furthermore, the original vote's margin of defeat was <math>(400-a)-a=400-2a</math>, and the revote's margin of victory was <math>b-(400-b)=2b-400</math>. Thus, from the problem, we know that <math>2(400-2a)=2b-400</math>. Substituting the value of <math>b</math> we obtained earlier into this equation, we see that: | ||
| + | \begin{align*} \\ | ||
| + | 2(400-2a)&=2*\frac{12}{11}(400-a)-400 \\ | ||
| + | 400-2a&=\frac{400*12}{11}-\frac{12a}{11}-200 \\ | ||
| + | 200-a&=\frac{400*6}{11}-\frac{6a}{11}-100 \\ | ||
| + | 300-\frac{2400}{11}&=\frac{5a}{11} \\ | ||
| + | 5a&=3300-2400 \\ | ||
| + | 5a&=900 \\ | ||
| + | a&=180 \\ | ||
| + | \end{align*} | ||
| + | From this fact, we can conclude that <math>c=\frac{12}{11}*(400-180)=240</math>, so <math>c-a=\boxed{60}</math>, which is answer choice <math>\fbox{B}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 08:53, 18 July 2024
Problem
With 
members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was 
of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?
Solution
Let the number of votes for the bill the first time be 
and the number of votes for the bill the second time be 
. Then, from the problem, we know that 
. Furthermore, the original vote's margin of defeat was 
, and the revote's margin of victory was 
. Thus, from the problem, we know that 
. Substituting the value of we obtained earlier into this equation, we see that:
\begin{align*} \\
2(400-2a)&=2*\frac{12}{11}(400-a)-400 \\
400-2a&=\frac{400*12}{11}-\frac{12a}{11}-200 \\
200-a&=\frac{400*6}{11}-\frac{6a}{11}-100 \\
300-\frac{2400}{11}&=\frac{5a}{11} \\
5a&=3300-2400 \\
5a&=900 \\
a&=180 \\
\end{align*}
From this fact, we can conclude that 
, so 
, which is answer choice 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
